Heine–Cantor theorem

Statement

Let $X$ and $Y$ be metric spaces and let $f \colon X \to Y$ be a continuous map. If $X$ is compact, then $f$ is uniformly continuous.

Remarks

Proof by contradiction: if $f$ is not uniformly continuous, there are $\varepsilon > 0$ and points $x_n, y_n$ with $d(x_n, y_n) \to 0$ and $d(f(x_n), f(y_n)) \ge \varepsilon$. By sequential compactness, a subsequence of $(x_n)$ converges to some $x$, and the matching subsequence of $(y_n)$ also converges to $x$. Continuity at $x$ gives a contradiction. For example, each continuous $f \colon [a, b] \to \mathbb{R}$ is uniformly continuous. This fact is used to show that $f$ is Riemann integrable. The theorem fails on $\mathbb{R}$: $x \mapsto x^2$ is continuous but not uniformly continuous.

Explanations

L1 human

On a compact space, continuity at each point is enough to get one $\delta$ for the whole space. Compactness rules out regions where the function becomes steeper and steeper without end.